nter Poi s
1
Conver t to Recur si ve Functi on
i nt sum(i nt[] nums, i nt si ze) { i nt sum = 0; for (i nt i ndex = 0; i ndex < si ze; i ndex++) { sum += nums[i ndex]; } r etur n sum; }
2
Poi nter s
Addi ti onal data type Poi nts to a l ocati on i n memor y Decl ar ati on:
i nt* x (poi nts a memor y l ocati on that contai ns i nt) doubl e* d (poi nts to mem l ocati on wi th a doubl e) Fr acti on* f (poi nt to mem. l ocati on wi th Fr acti on object)
Var i abl e and data ar e separ ated Getti ng poi nter to a var i abl e: & x Getti ng data poi nted to: * x
3
Exa mpl es
i nt x = 7; i nt* y = & x; *y = 9 What i s the val ue of x?
4
Ar r a ys a nd poi nter s
i nt
nums[10]
Type of nums: i nt[] or i nt* nums i s poi nti ng to the fi r st el ement
5
I ndexi ng i nto a n a r r a y Opti on 1
i nt sum(i nt* nums, i nt si ze) { i nt sum = 0; for (i nt i ndex = 0; i ndex < si ze; i ndex++) { sum = sum + nums[i ndex]; } r etur n sum; }
6
I ndexi ng i nto a n a r r a y Opti on 2
i nt sum(i nt* nums, i nt si ze) { i nt sum = 0; for (i nt count = 0; count < ze; si count++) { sum = sum + * nums; nums++; } r etur n sum; }
7
I ndexi ng i nto a n a r r a y Opti on 3
i nt sum(i nt* nums, i nt si ze) { i nt sum = 0; for (i nt i ndex = 0; i ndex < si ze; i ndex++) { sum = sum + * (nums + i ndex); } r etur n sum; }
8
H ea p ver sus Sta ck M emor y
Stack
M emor y
Stati c M ust know at compi l e ti me how much we'l l need L i nked di r ectl y wi th a var i abl e
H eap
M emor y
9
Dynami c Deter mi ne at r unti me how much needed Need poi nter to heap memor y Can't be named di r ectl y
Al l oca ti ng M emor y on the H ea p
i nt*
nums = new i nt[20]; i nt* nums = new i nt[n];
10
M a na gi...